Prove that the sum of eccentric angles of four concylic points on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is `2npi,` where ` n in Z`

Explanation

To prove that the sum of the eccentric angles of four concyclic points on the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(2n\pi\) where \(n \in \mathbb{Z}\), we can follow these steps:

### Step 1: Define the Points on the Ellipse
Let the four concyclic points on the ellipse be \(P, Q, R, S\). We can express the coordinates of these points in terms of their eccentric angles \(\theta_1, \theta_2, \theta_3, \theta_4\):
\[ P(a \cos \theta_1, b \sin \theta_1), \quad Q(a \cos \theta_2, b \sin \theta_2), \quad R(a \cos \theta_3, b \sin \theta_3), \quad S(a \cos \theta_4, b \sin \theta_4) \]

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